question_answer

The image of the point with position vector\[\mathbf{i}+3\mathbf{k}\]in the plane\[r\cdot (\mathbf{i}+\mathbf{j}+\mathbf{k})=1\]is

A) \[\mathbf{i}-2\mathbf{j}+\mathbf{k}\]               

B) \[\mathbf{i}+2\mathbf{j}-\mathbf{k}\]

C) \[-\mathbf{i}-2\mathbf{j}+\mathbf{k}\]              

D)         None of these

Correct Answer: C

Solution :

Let\[Q\]be the image of the point\[P(\mathbf{i}+3\mathbf{k})\]in the plane\[r(\mathbf{i}+\mathbf{j}+\mathbf{k})=1\]. Then, \[PQ\]is normal to the plane. Since,\[PQ\] passes through\[P\]and in normal to the given plane, therefore equation of\[PQ\]is\[\mathbf{r}=(\mathbf{i}+3\mathbf{k})+\lambda (\mathbf{i}+\mathbf{j}+\mathbf{k})\] Since,\[Q\]lies on the line\[PQ\], so let the position vector of\[Q\] be\[(\mathbf{i}+3\mathbf{k})+\lambda (\mathbf{i}+\mathbf{j}+\mathbf{k})\]i.e.,\[(1+\lambda )\mathbf{i}+\lambda \mathbf{j}+(3+\lambda )\mathbf{k}\]. Since,\[R\]is the mid-point of\[PQ\], therefore position vector of\[R\]is                 \[\frac{(1+\lambda )\mathbf{i}+\lambda \mathbf{j}+(3+\lambda )\mathbf{k}+\mathbf{i}+3\mathbf{k}}{2}\]                 \[=\left( \frac{\lambda +2}{2} \right)\mathbf{i}+\frac{\lambda }{2}\mathbf{j}+\left( \frac{6+\lambda }{2} \right)\mathbf{k}\]                 \[=\left( \frac{\lambda }{2}+1 \right)\mathbf{i}+\frac{\lambda }{2}\mathbf{j}+\left( 3+\frac{\lambda }{2} \right)\mathbf{k}\] Since,\[R\]lies on the plane                 \[r.(i+j+k)=1\] \[\therefore \,\,\,\,\,\left[ \left( \frac{\lambda }{2}+1 \right)i+\frac{\lambda }{2}j+\left( 3+\frac{\lambda }{2} \right)k \right]\]                 \[.(i+j+k)=1\] \[\Rightarrow \,\,\,\left[ \frac{\lambda }{2}+1+\frac{\lambda }{2}+3+\frac{\lambda }{2} \right]=1\] \[\Rightarrow \,\,\lambda =-2\] Hence, the position vector of Q is \[-i-2j+k\].